# How do you solve 3^ { 3r - 1} = 3^ { - 2r - 3}?

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#### Explanation

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Jim G. Share
Apr 16, 2017

$r = - \frac{2}{5}$

#### Explanation:

Since the $\textcolor{b l u e}{\text{bases on both sides of the equation are 3}}$ we can equate the exponents.

$\Rightarrow 3 r - 1 = - 2 r - 3$

$\Rightarrow 5 r = - 2$

$\Rightarrow r = - \frac{2}{5}$

$\textcolor{b l u e}{\text{As a check}}$

${3}^{\left(- \frac{6}{5} - \frac{5}{5}\right)} = {3}^{- \frac{11}{5}}$

$\text{and } {3}^{\left(\frac{4}{5} - \frac{15}{5}\right)} = {3}^{- \frac{11}{5}}$

$\Rightarrow r = - \frac{2}{5} \text{ is the solution}$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Apr 16, 2017

In the given eqn, the bases are equal, so equate powers on both sides.

#### Explanation:

$3 r - 1 = - 2 r - 3$

$5 r = - 2$

$r = - \frac{2}{5}$

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